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Let's solve the given problems by factoring the differences of cubes.

### The difference of cubes formula is:
\[
a^3 - b^3 = (a - b)(a^2 + ab + b^2)
\]

### **Problem a) \( 64x^3 - 27 \)**

1. First, recognize that both terms are cubes:
   \[
   64x^3 = (4x)^3 \quad \text{and} \quad 27 = 3^3
   \]
2. Apply the difference of cubes formula:
   \[
   (4x)^3 - 3^3 = (4x - 3)\left((4x)^2 + (4x)(3) + 3^2\right)
   \]
3. Simplify the second factor:
   \[
   = (4x - 3)\left(16x^2 + 12x + 9\right)
   \]
   
So, the factored form of \( 64x^3 - 27 \) is:
\[
(4x - 3)(16x^2 + 12x + 9)
\]

---

### **Problem c) \( 343x^3 - 1728 \)**

1. Again, recognize that both terms are cubes:
   \[
   343x^3 = (7x)^3 \quad \text{and} \quad 1728 = 12^3
   \]
2. Apply the difference of cubes formula:
   \[
   (7x)^3 - 12^3 = (7x - 12)\left((7x)^2 + (7x)(12) + 12^2\right)
   \]
3. Simplify the second factor:
   \[
   = (7x - 12)\left(49x^2 + 84x + 144\right)
   \]

So, the factored form of \( 343x^3 - 1728 \) is:
\[
(7x - 12)(49x^2 + 84x + 144)
\]

Would you like further clarification or details on these factorizations? 

Here are some related questions you might want to explore:
1. How do you factor the sum of cubes?
2. What happens when factoring higher powers, like quartic polynomials?
3. Can all cubic polynomials be factored similarly?
4. What real-world applications involve factoring polynomials?
5. How do you solve equations involving cubic expressions?

**Tip:** When factoring cubes, always verify whether both terms are perfect cubes before applying the formula!

Factor each difference of cubes: 64x^3 - 27 and 343x^3 - 1728

Learn how to factor the difference of cubes for polynomials like 64x^3 - 27 and 343x^3 - 1728 using the difference of cubes formula. Suitable for Grades 9-11.

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